I’ve been told that a greater binding energy means the nucleus is more tightly bound, and therefore that decreases the mass of the nucleus with respect to its nucleons when separated. But why does a tighter nucleus mean a smaller mass?

  • $\begingroup$ The question seems to be circular. You explained why. Unless you're struggling with the idea that mass is a kind of energy is mass which is simply a feature of the universe as explained in relativity. $\endgroup$ Jul 14 '15 at 14:38

The force between nucleons is rather complicated, so let's consider the simpler example of assembling a hydrogen atom from a proton and an electron.

We start with the proton and electron at rest and a long way apart. The kinetic energy is zero (because they're at rest) and the potential energy is given by:

$$ V = -k\frac{Q_1Q_2}{r} \tag{1} $$

where $Q_1$ and $Q_2$ are both the electron charge $e$, and $r$ is the separation. Since we start with the two particles far apart, i.e. $r \rightarrow \infty$, the potential energy is also (approximately) zero. The only energy present is due to the rest masses of the proton and electron i.e.

$$ E = m_pc^2 + m_ec^2 $$

Now we let the electron and proton move towards each other due to the electrostatic force between them. We aren't adding any energy, so the total energy must stay the same, but the potential energy (given by equation 1) is decreasing as $r$ decreases, so the kinetic energy must increase to balance it. To make life easy we'll assume the proton stays stationary and only the electron moves - the proton is so much heavier than the electron that this isn't a bad approximation - in which case the kinetic energy of the electron is:

$$ \tfrac{1}{2}m_ev^2 = k\frac{e^2}{r} \tag{2} $$

So far so good, but here's the problem. In a hydrogen atom the separation of the electron and proton is about a Bohr radius, which is about 0.05 nm. If we let the electron fall to one Bohr radius from the proton then use equation (2) to calculate the kinetic energy of the electron we find it is about 27.2 eV, which corresponds to a velocity of about 3 million metres per second. The electron is moving far too fast to "stick" to the proton. It will just flash past the proton and whizz off to infinity again.

To make the hydrogen atom we have to slow down the electron, that is we have to take away some of its kinetic energy (in fact we have to take away about 13.6 eV of its energy). This will slow the electron down enough to make it "stick" (I'm using the term "stick" rather loosely here!) to the proton.

Now, we started with an energy:

$$ E = m_pc^2 + m_ec^2 $$

and we have to take away 13.6 eV or our hydrogen atom won't form. So the energy of the hydrogen atom is:

$$ E_H = m_pc^2 + m_ec^2 - 13.6 eV $$

The mass of the hydrogen atom is given by Einstein's famous equation $E = mc^2$, so to get the mass of the hydrogen atom we divide by $c^2$ to get:

$$ m_H = m_p + m_e - \frac{13.6 eV}{c^2} $$

And you can see immediately that the mass of the hydrogen atom is less than the mass of the proton and electron we used to make it. The mass is less because we started with a proton and electron then took something away. In practice when we combine protons and electrons they form hydrogen by emitting a photon with the energy 13.6 eV.

The argument is the same for nuclei. Nuclei are fiendishly complicated beasts, but if we imagine making a nucleus by starting with a bunch of protons and neutrons and letting them fall together, then we have to take away some of their kinetic energy to make them stick. If we divide the energy we took away by $c^2$ this tells us the mass we took away i.e. how much lighter the nucleus is than the protons and neutrons we started with.

  • $\begingroup$ @John_Rennie: I noticed that the 13.6eV to "capture" the electron is exactly $\frac{1}{2}$ of the 27.2eV of kinetic energy of the electron whizzing past the proton. Is this more than just a coincidence? $\endgroup$
    – pr1268
    Aug 17 '16 at 2:07
  • 1
    $\begingroup$ @pr1268: indeed not, well spotted. This is a result of the virial theorem. $\endgroup$ Aug 17 '16 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.